WOOD, SOILS, AND STEEL INTRO

KNOWLEDGE BASE REQUIRED:

STRENGTH OF MATERIALS

STEEL DESIGN

SOIL MECHANICS

REVIEW OF TIMBER DESIGN

•BENDING MEMBERS•DEFLECTION MEMBERS•SHEAR MEMBERS•COLUMN MEMBER •BEARING PROBLEM

TIMBER DESIGN

BEARING PERPENDICULAR TO THE GRAIN- fc(perp)

REVIEW OF TIMBER:

lb+3/8

lbcpercper F

AP

f '' P

where lb= bearing length

Note: When the bearing length is less than 6 in. and when the distance from the end of the beam to the contact area is more than 3 in., the allowable bearing stress may be increased by Cb.

b

bb l

lC

375.0

The deformation limit of .04 inch. is provided by ASTM D143 provides adequate service in typical wood-frame construction.

Special Cases In some designs where the deformation is critical, a reduced value can be applied. ( WWPA P.9 Table F)

Deflection can be designed for a reduce limit of .02 in. (also refer to P.251 in text)

Fc (perp .02) = 0.73 Fc (perp .04) + 5.60

Sample Problem:

Given a Hem-Fir Select Structural with 11,000#s on supports:a) check for the bearing of a cantilever support. b) Assume critical deflection for heavy impact loads at end of cantilever.

2 - 2x12

4x8

> 3”

3.5”

1.5” 1.5”

Fc(perp) = 405 psilb= 3”therefore we can increase bearing stress, but letsbe conservative and use lb as recommended

psipsixx

fcper 40510475.35.12

#000,11 N.G.

We have to increase bearing

V 11,000Req’d Area= ----- = ---------- = 27 sq in. Fc(perp) 405psi

add2-2X12 X 12 A= 6 X 3.5 = 21 sq in < 27 sq in NG 2-3X12 X 12 A=[(2X1.5)+(2X2.5)](3.5)= 28 sq in > 27 sq in OK

b) 4x8 bearing problem is O.K., now solve for critical deflection with limit of .02 inch

F’c(perp .02) = 0.73 (405) + 5.60F’c(perp .02)= 301.25 psi

Req’d Area = 11000/301.25 =36.5 sq in

add 2- 4x12x12 A=[(2X3.5)+(2X1.5)](3.5)= 35 sq in N.G.

use 2- 6x12 - 49 sq in. or a steel plate 3.5X10.5

REVIEW OF SOIL MECHANICS

•VERTICAL STRESSES• LATERAL STRESSES

LECTURE #4

BASIC SOIL MECHANICS REVIEW:

= UNIT WEIGHT OF SOIL (PCF, KN/m3)

sat = SATURATED UNIT WEIGHT OF SOIL

b = BOUYANT UNIT WEIGHT OF SOIL

W = UNIT WEIGHT OF WATER(62.4PCF,9.81 KN/m3)

LECTURE #4(CONT)

BASIC SOIL MECHANICS REVIEW:

satb W= -

VERTICAL STRESSES:

V = VERTICAL STRESS (PSF, TSF,KN/m2)

hV

pcfsat 120

pcfdry 105G W T

3'

5 '

CALCULATE TOTAL AND EFFECTIVE V

(TOTAL STRESS)

Ka=0.5

WV h ' (EFFECTIVE STRESS)

LECTURE #4(CONT)

BASIC SOIL MECHANICS REVIEW:

VERTICAL STRESSES:

psfV

V

915600315

)5(120)3(105

)( hV

EFFECTIVE VERTICAL STRESSES:

)(' hV

psf

hh

V

V

bdV

603288315'

5)4.62120()3(105'

' 21

LECTURE #4(CONT)

BASIC SOIL MECHANICS REVIEW:

LATERAL STRESSES:

psfK

psfK

psfK

wVw

Vh

Vh

312)1)(5)(4.62(

144)5)(.288('

5.157)5)(.315(

22

1

Khh )(

LATERAL FORCE:

KhHF

hHF

H

V

)(5.0

)(5.0

LECTURE #4(CONT)

BASIC SOIL MECHANICS REVIEW:

LATERAL FORCES:

5' pcfsat 120

pcfdry 105G W T 157.5

144 312

3'

LECTURE #4(CONT)

BASIC SOIL MECHANICS REVIEW:

LATERAL FORCES:

5' pcfsat 120

pcfdry 105G W T 157.5

144 312

3'

TO FIND RESULTANT SUM FORCES: R

TO FIND RESULTANT LOCATION TAKE MOMENT:y

LECTURE #4(CONT)

BASIC SOIL MECHANICS REVIEW:

LATERAL FORCES:

5' pcfsat 120

pcfdry 105G W T 315

144312

3'

TO FIND RESULTANT SUM FORCES: R

TO FIND RESULTANT LOCATION TAKE MOMENT:y

236.25#

1575#

1140#

LECTURE #4(CONT)

BASIC SOIL MECHANICS REVIEW:

LATERAL FORCES:

R=236.25+1575+1140=2951.25#

.5.225.2951

)1140(3/5)1575(5.2)25.236(6

ftyRRy

y

LATERAL ARM:

LECTURE #4(CONT)

BASIC STEEL DESIGN REVIEW:

For temporary structures we use primarily involves:

Rolled Sections - W: sections Angle Sections Channel Sections

Very Rarely - Steel Joist Decking(thin gages) Trusses

Elements include Girders, Beams, Columns, & Struts (angles &channels)

LECTURE #4(CONT)

BASIC STEEL DESIGN REVIEW:

Review of AISC Construction Manual:

Allowable Stresses

Most commonly used is A36 SteelSteel Design Proceduresa) Use ASD - Allowable Stress Design Procedure(Basic Allowable Stress For A36 Steel) Tension- Ft=.6 Fy=14,400psi Shear - Fv= .4 Fy=32,400psi Bearing - Fp=.9Fy=32,400 psi Bending - Fb = .66Fy=23,760psiBased on Compact Section Compression=Fa (variable depending on unbraced length)

LECTURE #4(CONT)

BASIC STEEL DESIGN REVIEW:

Review of AISC Construction Manual (cont):

Compact vs. Non- CompactCompact Beam are rolled beam that can achievethe plastic moment.

non- com pact com pact

Stress Distribution of I Beam

LECTURE #4(CONT)

BASIC STEEL DESIGN REVIEW:

Review of AISC Construction Manual (cont):

Compact Sections-• are symmetrical about the y-y axis• webs/flanges must have certain web thickness ratios•compression flange must be adequately braced against lateral buckling

Design Procedure for Bending

1. Determine the Maximum Bending Moment2. Compute the Required Section Modulus based on allowable stresses Fb=.66Fy (compact) or Fb=.60 Fy (non compact)3.Lightest weight section is the most economical

LECTURE #4(CONT)

BASIC STEEL DESIGN REVIEW:

Review of AISC Construction Manual (cont):

Compression Members:• lc- distance between spacing of lateral bracesWhen brace spacing is less than or equal to lcthen Fb=.66Fy•Lu - maximum unsupported length When brace spacing is greater than Lc but less thanLu then Fb=.60Fy

When brace spacing is greater than lu, Fb is not determined and Total allowable moment has to be taken from a chart.